ECE 476

POWER SYSTEM ANALYSIS Lecture7

Development of Transmission Line Models Professor Tom Overbye Department of Electrical and Computer Engineering

Announcements

   For next two lectures read Chapter 5.

HW 2 is 4.10 (positive sequence is the same here as per phase), 4.18, 4.19, 4.23. Use Table A.4 values to determine the Geometric Mean Radius of the wires (i.e., the ninth column). Due September 15 in class. “Energy Tour” opportunity on Oct 1 from 9am to 9pm. Visit a coal power plant, a coal mine, a wind farm and a bio-diesel processing plant. Sponsored by Students for Environmental Concerns. Cost isn’t finalized, but should be between $10 and $20. Contact Rebecca Marcotte at [email protected]

for more information or to sign up.

1

SDGE Transmission Grid (From CALISO 2009 Transmission Plan)

2

Line Conductors

 Typical transmission lines use multi-strand conductors  ACSR (aluminum conductor steel reinforced) conductors are most common. A typical Al. to St. ratio is about 4 to 1.

3

Line Conductors, cont’d

  Total conductor area is given in circular mils. One circular mil is the area of a circle with a diameter of 0.001 =   0.0005

2 square inches

Example:

what is the the area of a solid, 1” diameter circular wire?

Answer:

1000 kcmil (kilo circular mils)  Because conductors are stranded, the equivalent radius must be provided by the manufacturer. In tables this value is known as the GMR and is usually expressed in feet.

4

Line Resistance

Line resistance per unit length is given by  R = A Example: What is the resistance in Ω / mile of a 1" diameter solid aluminum wire (at dc)?

R

    0.0127m

2 1609

m mile

  0.084

mile

5

Line Resistance, cont’d

   Because ac current tends to flow towards the surface of a conductor, the resistance of a line at 60 Hz is slightly higher than at dc.

Resistivity and hence line resistance increase as conductor temperature increases (changes is about 8% between 25  C and 50  C) Because ACSR conductors are stranded, actual resistance, inductance and capacitance needs to be determined from tables.

6

Variation in Line Resistance Example

7

Review of Electric Fields

To develop a model for line capacitance we first need to review some electric field concepts.

Gauss's law:  A = q e where (integrate over closed surface)

D

= electric flux density, coulombs/m 2 d = differential area da, with normal to surface A = total closed surface area, m 2

8

Gauss’s Law Example

Similar to Ampere’s Circuital law, Gauss’s Law is most useful for cases with symmetry.

Example: Calculate

D

about an infinitely long wire that has a charge density of q coulombs/meter.  A 

D

2 

Rh

 q e 

qh

Since

D

comes radially out inte grate over the cylinder bounding the wire

D

 2 

q R

a r

where

a r

radially directed unit vector

9

Electric Fields

The electric field,

E

, is related to the electric flux density,

D

, by

D

= 

E

where

E

   o  r = electric field (volts/m) = permittivity in farads/m (F/m) =  o  r = permittivity of free space (8.854

 10 -12 F/m) = relative permittivity or the dielectric constant (  1 for dry air, 2 to 6 for most dielectrics)

10

Voltage Difference

The voltage difference between any two V    P  P  In previous example the voltage difference between  f  

o

) V     R R  

q

2 

o R dR



q

2 

o

ln

R



R



11

Voltage Difference, cont’d

With V     R R  

q

2 

o R dR



q

2 

o

ln

R



R

 if q is positive then those points closer in have a higher voltage. Voltage is defined as the energy (in Joules) required to move a 1 coulomb charge against an ele ctric field (Joules/Coulomb). Voltage is infinite if we pick infinity as the reference point

12

Multi-Conductor Case

Now assume we have n parallel conductors, The voltage difference between our two points, V   1 2 

i n

  1

q i

ln

R



R



i i

where

R



i

is the radial distance from point P  to conductor i, and

R



i

13

Multi Conductor Case, cont’d

If we assume that n  i=1 q i  0 then rewriting V   1 2 

i n

  1

q i

1 ln

R



i

 1 2 

i n

  1

q i

ln

R



i

We then subtract

i n

  1

q i

ln

R

 1  0 V   1 2 

i n

  1

q i

1 ln

R



i

 1 2 

i n

  1

q i

ln

R



R

 1

i R



i R

 1  0

14

Absolute Voltage Defined

infinity, we can now define the voltage of a point w.r.t. a reference voltage at infinity: V   1 2 

i n

  1

q i

1 ln

R



i

This equation holds for any point as long a s it is not inside one of the wires!

15

Three Conductor Case

C A

V a



V a

 1 2  B Assume we have three infinitely long conductors, A, B, & C, each with radius r and distance D from the other two conductors.  

q a

Assume charge densities such ln that q a + q b + q c = 0 1  ln 1 ln 1

r q b D



q c D

 

q a

2  ln

D r

16

Line Capacitance

For a single line capacitance is defined as

q i

 But for a multiple conductor case we need to use matrix relationships since the charge on conductor i may be a function of V j

q

1     

q



C V

   

C

11

C n

1

C

1

n C nn V

1         

17

Line Capacitance, cont’d

In ECE 476 we will not be considering theses cases with mutual capacitance. To eliminate mutual capacitance we'll again assume we have a uniformly transposed line. For the previous three conductor exam ple:

V a



V S V

a 

C

 q a

V a

 2  ln

D r

18

Bundled Conductor Capacitance

Similar to what we did for determining line inductance when there are n bundled conductors, we use the original capacitance equation just substituting an equivalent r adius R c b  (

rd

12

d

1

n

) 1

n

Note fo r t he capacitance equation we use r rather equation

19

Line Capacitance, cont’d

For the case of uniformly transposed lines we

C

 2  ln

D m R c b

where D m    1 3 R c b  (

rd

12

d

1

n

) 1

n

(note r NOT r')  o  

20

Line Capacitance Example

Calculate the per phase capacitance and susceptance of a balanced 3  , 60 Hz, transmission line with horizontal phase spacing of 10m using three conductor bundling with a spacing between conductors in the bundle of 0.3m. Assume the line is uniformly transposed and the conductors have a a 1cm radius.

21

Line Capacitance Example, cont’d

R c b

D m

C

X c      1 3  0.0963 m 2   12.6

ln 0.0963

1 3   12 1 

C

    12.6 m   11 F/m 1  / m)  11 F/m

22

ACSR Table Data (Similar to Table A.4)

GMR is equivalent to r’ Inductance and Capacitance assume a D m of 1 ft.

23

ACSR Data, cont’d

X L    2 

f L

 3

f

 4 

f

  ln 1

GMR

 10  7 ln

D m GMR

  ln

D m

   3

f

1 ln

GMR

  3

f

ln

D m

Term from table assuming a one foot spacing Term independent of conductor with D m in feet .

24

ACSR Data, Cont.

To use the phase to neutral capacitance from table X C  2  1

f C

 -m where

C

 2  0 ln

D m r



f

1  

D m r



f

1  1

r f D m

Term from table assuming a one foot spacing Term independent of conductor with D m in feet .

25

Dove Example

GMR

 0.0313 feet Outside Diameter = 0.07725 feet (radius = 0.03863) Assuming a one foot spacing at 60 Hz

X a X a

   7  1 0.0313

Ω/mile  0.420 Ω/mile, which matches the table For the capacitance

X C

 1 

f

1 

r

26

Additional Transmission Topics



Multi-circuit lines

: Multiple lines often share a common transmission right-of-way. This DOES cause mutual inductance and capacitance, but is often ignored in system analysis. 

Cables:

There are about 3000 miles of underground ac cables in U.S. Cables are primarily used in urban areas. In a cable the conductors are tightly spaced, (< 1ft) with oil impregnated paper commonly used to provide insulation – inductance is lower – capacitance is higher, limiting cable length

27

Additional Transmission topics



Ground wires:

Transmission lines are usually protected from lightning strikes with a ground wire. This topmost wire (or wires) helps to attenuate the transient voltages/currents that arise during a lighting strike. The ground wire is typically grounded at each pole. 

Corona discharge:

Due to high electric fields around lines, the air molecules become ionized. This causes a crackling sound and may cause the line to glow!

28

Additional Transmission topics



Shunt conductance:

Usually ignored. A small current may flow through contaminants on insulators.



DC Transmission:

Because of the large fixed cost necessary to convert ac to dc and then back to ac, dc transmission is only practical for several specialized applications – long distance overhead power transfer (> 400 miles) – long cable power transfer such as underwater – providing an asynchronous means of joining different power systems (such as the Eastern and Western grids).

29

Tree Trimming: Before

30

Tree Trimming: After

31

Transmission Line Models

 Previous lectures have covered how to calculate the distributed inductance, capacitance and resistance of transmission lines.

 In this section we will use these distributed parameters to develop the transmission line models used in power system analysis.

32

Transmission Line Equivalent Circuit

Our current model of a transmission line is shown below Units on z and y are per unit length!

33

Derivation of V, I Relationships

We can then derive the following relationships:

dV



I z dx dI

 (

V

 

V y dx



z I dx dx



yV

34

Setting up a Second Order Equation



z I dx



yV dx

We can rewrite these two, first order differential equations as a single second order equation ( ) 

z dx

2

dx

( ) 

dx

2

zyV

 0 

zyV

35

V, I Relationships, cont’d

  where    

yz

  

j

 the attenuation constant the phase constant Use the Laplace Transform to solve. System has a characteristic equation (

s

2   2 )

s

 )(

s

  )  0

36

Equation for Voltage

The general equation for V is  

x

  

x

Which can be rewritten as  (

k

1 

k

2 )(

e



x

 2

e

 

x

)  (

k

1 

k

2 )(

e



x



e

 

x

) 2 Let K 1  

K

1

K

1

k

2 and K 2

e



x

 ( cosh(  2

e

 

x

) 

x

) 

K

2

K

2 (

k

2 . Then

e



x



e

 

x

2 sinh( 

x

) )

37

Real Hyperbolic Functions

For real x the cosh and sinh functions have the following form:

d

cosh( 

x

)   sinh( 

x

)

dx d

sinh( 

x

)

dx

  cosh( 

x

)

38

Complex Hyperbolic Functions

For x =  + j  the cosh and sinh functions have the following form cosh

x

sinh

x

     

j j

 

39